Problem: This is the easy version of the problem. The difference between the versions is that in the easy version all prices a_i are different. You can make hacks if and only if you solved both versions of the problem.

Today is Sage's birthday, and she will go shopping to buy ice spheres. All n ice spheres are placed in a row and they are numbered from 1 to n from left to right. Each ice sphere has a positive integer price. In this version all prices are different.

An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.

You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of ice spheres in the shop.

The second line contains n different integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the prices of ice spheres.

Output

In the first line print the maximum number of ice spheres that Sage can buy.

In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.

Example

Input


5
1 2 3 4 5


Output


2
3 1 4 2 5 

Note

In the example it's not possible to place ice spheres in any order so that Sage would buy 3 of them. If the ice spheres are placed like this (3, 1, 4, 2, 5), then Sage will buy two spheres: one for 1 and one for 2, because they are cheap.

Solution (in PYTHON3):

Solution:
## necessary imports
import sys
input = sys.stdin.readline
from bisect import bisect_left;
from bisect import bisect_right;
from math import ceil, factorial;

def ceil(x):
    if x != int(x):
        x = int(x) + 1;
    return x;
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;

## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);

## nCr function efficient using Binomial Cofficient
def nCr(n, k, modulus = 1): 
    if(k > n - k): 
        k = n - k; 
    res = 1;
    for i in range(k): 
        res = res * (n - i);
        res = res / (i + 1);
        res %= modulus; 
    return int(res);

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret;
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));

def float_array():
    return list(map(float, input().strip().split()));

## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

n = int(input()); a = sorted(int_array());
if n < 3:
    print(0); print(*a);
else:
    ans = (n - 1) // 2;
    i = 0;
    while(i < n - 1):
        temp = a[i];
        a[i] = a[i + 1];
        a[i + 1] = temp;
        i += 2;
    print(ans); print(*a);